Question

A signal of \[0.1\,KW\] is transmitted in a cable. The attenuation of cable is \[ - 5\,dB\] per km and cable length is \[20\,km\]. The power received at the receiver is \[{10^{ - x}}W\]. The value of \[x\] is…\[\left( {gain\,\,in\,\,dB = \left[ {10{{\log }_{10}}\left( {\dfrac{{{p_0}}}{{{p_i}}}} \right)} \right]} \right)\]

Answer 1

Hint:We know that Loss of signal intensity in networking cables or connections is referred to as attenuation in networking or signal communication. In this question, we need to substitute the given value in the gain formula to find the value of \[x\].

Formula used:
We need to use the following formula:
1. \[1\,\,KW = 1000\,\,W\]
2. \[\log \,{x^n} = n \cdot \log \,x\]
3. \[\log \,10 = 1\]

Complete step by step solution:
We are given that,
\[\left( {gain\,\,in\,\,dB = \left[ {10{{\log }_{10}}\left( {\dfrac{{{p_0}}}{{{p_i}}}} \right)} \right]} \right)...\left( 1 \right)\]
Now the power of the signal transmitted \[{p_i} = 0.1\,KW\].
Now we are asked to find the power received at the receiver in Watt
So, \[{p_i} = 0.1\, \times {10^3}\,W(\because 1KW = {10^3}W)\]
Now the rate of attenuation \[ = - 5dB/Km\]
And the total length of a path \[ = 20\,Km\]
Thus, the total loss suffered \[ = - 5 \times 20\,\,dB\]

Now by substituting all the values in equation (1), we get
\[\text{Gain in dB} = 10\,{\log _{10}}\left( {\dfrac{{{p_0}}}{{{p_i}}}} \right) \\
\Rightarrow - 100 - 0 = 10{\log _{10}}\left( {\dfrac{{{p_0}}}{{{p_i}}}} \right) \\
\Rightarrow - 100 = 10{\log _{10}}\left( {\dfrac{{{p_0}}}{{{p_i}}}} \right) \\
\Rightarrow 100 = 10{\log _{10}}\left( {\dfrac{{{p_i}}}{{{p_0}}}} \right) \\ \]
Further solving we get,
\[{\log _{10}}\left( {\dfrac{{{p_i}}}{{{p_0}}}} \right) = \dfrac{{100}}{{10}} \\
\Rightarrow {\log _{10}}\left( {\dfrac{{{p_i}}}{{{p_0}}}} \right) = 10 \\
\Rightarrow {\log _{10}}\left( {\dfrac{{{p_i}}}{{{p_0}}}} \right) = {\log _{10}}{\left( {10} \right)^{10}}\left( {\because {{\log }_{10}}{{\left( {10} \right)}^{10}} = 10} \right) \\ \]
Now by canceling log on both sides, we get
\[\left( {\dfrac{{{p_i}}}{{{p_0}}}} \right) = {\left( {10} \right)^{10}}\]

Now by substituting the value of \[{p_i}\], we get
\[\dfrac{{1.0 \times {{10}^3}}}{{{p_0}}} = {10^{10}} \\
\Rightarrow 1.0 \times {10^3} = {10^{10}} \times {p_0} \\
\Rightarrow {p_0} = 1.0 \times {10^3} \times {10^{ - 10}}\,W \\
\Rightarrow {p_0} = 1.0 \times {10^{ - 7}}\,W \\ \]
Further solving we get,
\[{p_0} = 1 \times {10^{ - 8}}\,W \\
\therefore {p_0} = {10^{ - 8}}\,W \]
Therefore, the power received at the receiver is \[{10^{ - 8}}\,W\].

Hence, the value of \[\,x\] is \[\,8\].

Additional information: The loss between the provider at the station and the receiver at home is measured by line attenuation on a digital subscriber line.

Note: Students calculated while replacing variables and became perplexed by the total loss formula. So, for the purpose of convenience, recall all formulas correctly so that they do not have any problems in the future.